Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 4 / 5F (Numbers and Sets)

The recurrence and the addition formula give the result by induction. It is true for $n=0,1$, and if true for $n,n-1$, then

The recurrence shows that $T_n$ is a polynomial with integer coefficients. Part (a) givesThus $\cos (\pi /n)$ is a root of a nonzero integer polynomial and is algebraic.

Put $z=2\cos (\pi /n)$. The recurrencedefines monic integer polynomials and gives $S_k(z)=2\cos (k\pi /n)$. Hence $z$ is an algebraic integer. If $\cos (\pi /n)$ is rational, then the rational algebraic integer $z$ is an integer. Since $-2\le z<2$, this leaves the values corresponding toThis is the rational cosine of an integral submultiple of pi result. For every $n\ge 4$, the value lies strictly between $1/2$ and $1$, so it is irrational.

Every term is nonnegative. Since $1-\cos x\le x^2/2$,Comparison with $\sum k^{-2}$ shows that the partial sums are bounded. Being increasing, they converge to the finite limit

Taking real parts of the geometric sum givesso $(a_n)$ is bounded when $\cos y\ne 1$. It cannot converge: convergence would imply $\cos (ny)=a_n-a_{n-1}\to 0$, but thenwhereas the same necessary condition applied to the subsequence $2n$ would give a limit of zero.